text:
How is varying modulation depth achieved by localizer ground transmitters?
An approaching aircraft's navigation receiver uses the difference in depth of modulation between the 90Hz and 150Hz signals which are amplitude modulated onto an RF carrier (on a channel between 108Mhz and 118MHz).
My understanding is that the ground transmitter is formed by several pairs of transmitters forming two interleaved phased arrays to achieved directional beam forming.
My questions is how is the difference in modulation depth achieved by the transmitter system based upon aircraft's position relative to the landing strip's entreline? The beam strength decreases as you move away from it's own centreline, so is it actually that the entire modulated signal strength decreases which when de-modulated is effectively a difference in amplitude modulation depth?

Possible duplicate of How does ILS (Instrument landing system) work?

The 90 Hz and 150 Hz beams are separate beams and the Localizer path is the 'valley' in the middle.

The other question has an answer which says "The position of the needle is determined by which lobe of the transmitter is being received stronger than the other", so it doesn't answer the depth of modulation part (amplitude is not modulation depth).

The beam strength decreases as you move away from it's own centreline,
so is it actually that the entire modulated signal strength decreases which
when de-modulated is effectively a difference in amplitude modulation depth?

Your question is a really good one. The depth of modulation is the ratio between the modulation amplitude and the carrier amplitude. Changing the relative strength of a signal changes both amplitudes in the ratio, but keeps the ratio unchanged. No DDM can indeed appear this way.
The 90 and 150 Hz tones are not used directly. The heart of the system creating the DDM is two signals: CSB (carrier and sidebands, see below what is a sideband) and SBO (sidebands only) and surprise! they are containing the same quantity of 90 and 150 Hz tones and are transmitted in equal quantity on both sides of the runway.
Now the key points of ILS:

SBO phase depends on the runway side (+/-90°, CSB is kept unchanged)
Phase of the 150 Hz tone is inverted in SBO (90 Hz is kept unchanged).

As SBO contains only the sidebands when CSB and SBO interfere in space, only the sidebands can actually interfere, the carrier found in CSB is unaltered.
When the sidebands interfere, the way they are constructed (by tweaking phases) makes 150 Hz more prominent on the left of the runway, and less prominent on the right side (details below, drawings will clarify). So:

The carrier amplitude hasn't changed in the process
The sidebands amplitude has been altered.

As the sidebands are the modulation, in effect there is a change in the DM (the ratio between modulation and carrier amplitude), and a DDM appears magically, increasing as we move away from CSB, that is as we move away from the centerline.
Countless oversimplified descriptions put the focus on 90 and 150 Hz, each one allegedly transmitted on a specific side of the runway (cf. image above). They miss the point the ILS is not based on amplitude comparison, but on depth of modulation comparison.
Two last points:

A precise location cannot be determined by simply using received signal strength (RSS) of two or more signals. RSS-based positioning improved by averaging algorithms is commonly found for rough positioning (e.g. in cellular networks) because the cost is marginal, but RSS depends on many factors which cannot be controlled.

On the other hand, the phase of a signal remains unaltered until the signal is reflected, diffracted or enters a material with a different refractive index. By taking appropriate protection measures in the Fresnel zone, the phase-based ILS is therefore very reliable. Phase is used in many sophisticated radio applications. Missing the phase role in the ILS (and conventional VOR) is actually missing the cleverness of these technologies.

If this summary increased your curiosity for more details, just follow on, but for a good understanding, we'll need to first clarify a few notions: Phase, modulation and sidebands.
Phase and phase comparison
When representing a signal, e.g. a sine signal, we can look at different parameters. Often we draw the amplitude as a function of the time (right side):

Two representations of a signal: Phase/amplitude and time/amplitude
We can also imagine a circle with a scale from 0° to 360°, a point traveling on this scale in the time of a signal cycle (left side above).
The phase is the value of the angle. We are usually interested in the phase difference of two signals with the same frequency:

Phase difference
The two signals can be in phase (no difference), in phase opposition (180°), in quadrature (90°) or have any phase difference. The phase difference plays a major role when two signals interfere:

If the signal are in phase, the amplitude of the result is the sum of the individual amplitudes.
If the signals are in phase opposition, the amplitudes subtract.
For any another phase difference, the amplitude of the result is between zero and the sum of individual amplitudes.

In the ILS, basically one signal is sent for reference (CSB) and one signal is sent for comparison (SBO). SBO on one side creates additive interferences with CSB, and substractive interferences on the other side. But actually because 150 Hz is inverted in SBO, on the additive side, at the center of the interference area, 150 Hz cancels itself, 90 Hz doubles its amplitude. On the other side 150 Hz doubles its amplitude, 90 Hz is cancelled. As we move away from the center of the interference area, this additive/subtractive effect decreases until SBO is left unaltered because we are out of CSB area.
Now let's clarify the part related to the modulation and depth of modulation.
Amplitude modulation, depth of modulation, sidebands
Modulation relates to a carrier, which is just a fancy name for a constant (usually sine) RF signal at a given frequency, therefore with all the wave energy concentrated at this frequency. The frequency of the carrier is usually very large compared to the modulation.
The modulating signal (or modulation) is the signal to transmit, it can have any shape, but in our case its shape is a 90 Hz or 150 Hz sine curve.
The amplitude modulation consists in changing the carrier instantaneous amplitude according to the modulation instantaneous amplitude. After it's done, the  carrier outer envelope reflects the modulation outer envelope:

Carrier modulated at 50%
The modulation can imprint itself at a variable depth (taller or smaller peaks). This is measured by the modulation index aka depth of modulation. The index value is an amplitude ratio: modulation amplitude / carrier amplitude.
The depth of modulation used for the LOC signal is only 20% (40% for the G/S signal) to leave some room for identification tones on 1020 Hz.
ICAO definition for ILS modulation depth:

The depth of modulation is the ratio of the amplitude of the modulation of
the 90 Hz or 150 Hz signal to the carrier amplitude. The DDM is the
modulation depth of the stronger signal minus the modulation depth of the
weaker signal.

Let's look at how energy is distributed according to frequency.

Spectrum of a carrier AM modulated by a sine tone
Before  modulation, all energy is concentrated at the carrier frequency (blue bar below). The modulation process creates two variable sidebands: lower (LSB) and upper (USB), depicted in magenta.
If the modulation index is 100%, the carrier contains 66% of the energy and each sideband 17% (mathematical explanation). When a 100 MHz carrier is modulated by a 150 Hz tone, the LSB is created at 100 MHz - 150 Hz, and the USB at 100 MHz + 150 Hz.
The larger the depth of modulation, the larger the energy transferred from the carrier to the  sidebands and the lower the energy remaining at the carrier frequency. Sideband energy and depth of modulation are the two sides of the same coin. In the next  paragraphs I'll talk about energy rather than depth, but it's the same.
Modulating the ILS signal with 90/150 Hz tones
The localizer signal is modulated by two tones, the reference at 90 Hz, the signal to compare at 150 Hz. Therefore each sideband energy is split between two  frequencies:

CSB signal spectrum
This signal, known as CSB (carrier and sidebands) will be used as-is in the localizer. However as explained earlier, the in-phase / out-of-phase element must be sent along with CSB.
From 90/150 Hz modulations to CSB/SBO modulations
The localizer transmitter sends an additional signal with an inverted 150 Hz, known as SBO (sidebands only). For this signal the carrier element is cancelled (DSB-SC modulation):

SBO signal spectrum with 150 Hz modulation inverted
The signal is sent in phase opposition with CSB on the left side of the runway, and in phase with CSB on the right side (this is done using phase shifters at the antenna level, see further down).
To summarize:

The localizer transmits CSB on the left and right sides of the array. CSB contains both 150 Hz and 90 Hz with no phase adjustment, and the carrier.

The localizer also transmits SBO which contains only the sidebands, but where the 150 Hz modulation is phase-shifted by 180° on both sides. The SBO signal itself is sent with a different phase on each half of the antenna array.

How CSB and SBO are sent by the antennas
The actual phase between SBO and CSB is adjusted to compensate for space modulation effects, and SBO is sent with a +90° phase relative to CSB for the antennas on the left of the runway centerline, and with a -90° phase relative to CSB for the antennas of the right of the enterline (the two sides of the array still have a 180° phase difference).
The combination of exact opposite phases for SBO signal cancels the EM field at the center, splitting the SBO radiation pattern in two portions, it seems there are two beams for SBO. CSB is radiated centered on this hole:

CSB + SBO radiation pattern
The elements compositing CSB and SBO must be synchrone. Phases are adjusted carefully to form the radiation pattern (i.e. align the loalizer with the centerline and set its width) and to ensure the linearity of the DDM in the central portion.
When spread in space, the three signals mix by space modulation, the principle the ILS is based on and which is more simply known as wave interference.
Space modulation
One signal (SBO) has its carrier suppressed preventing the carrier energy to be increased during the mix CSB+SBO. In contrary when mixing the sidebands from each signal, the space modulation adds the sidebands energy where the sidebands are in phase, and subtract the  sidebands energy when they are in phase opposition.
SBO is in phase opposition with CSB on the left side of the runway and is in phase with CSB on the right side of the runway.

Along the centerline CSB only is present, with equal energy for 90 Hz (green) and 150 Hz (magenta). The LOC indicator will be centered. In order to create a CSB prominent area along the centerline, SBO power is also reduced on the center antennas.

On the left side, 150 Hz modulation has been shifted by 180° in SBO, and SBO signal has been shifted 180° relative to CSB at antenna level. 150 Hz modulation is therefore in phase for CSB and SBO. Sideband energies are added, while energy for 90 Hz modulation is  decreased because 90 Hz modulation in CSB and SBO are out of phase. So there is more energy for 150 Hz (which means there is a larger depth of modulation). This effect is getting stronger as we move away from the centerline.

On the right side where 150 Hz modulation is out of phase for CSB and SBO, 150 Hz energy is reduced compared to 90 Hz  energy in the sidebands, and the difference is getting stronger as we move away from the centerline.

Past 10° from the centerline, CSB cannot be sensed, only SBO can, the LOC indicator will deviate to full scale left or full scale right depending on the phase of  SBO. Another signal is used to ensure a full scale deviation past this point (see clearance signal further down).
If you want to do some space modulation simulation, here are the details of the signals to mix (see details here):

$E_{CSB} = E_C \space \cos \space 2 \pi f_C t + E_{90} \space \sin \space 2 \pi f_{90} t \space \cos \space 2 \pi f_C t + E_{150} \space \sin \space 2 \pi f_{150} t \space \cos \space 2 \pi F_C t$

$E_{SBO \space 90} = K \space \sin \space 2 \pi f_{90} t \space \cos \space 2 \pi f_C t$

$E_{SBO \space 150} = K \space \sin \space 2 \pi f_{150} t \space \cos \space 2 \pi f_C t$

We talked about the sidebands energy. Regarding the carrier, which is only present in CSB, the space modulation result always contains the energy of the original CSB  carrier until reaching full scale (then the clearance signal provides the carrier because while all modulation information is in the sidebands, the constant carrier  is necessary to synchronize and lock the receiver and to retrieve the coherent CSB and SBO from the modulated radiofrequency resulting signal).
Where the DDM finally appears
Now let wrap up:

The carrier energy is not changed by the space modulation as it is present in only one signal (CSB).
The sidebands energy is increased or decreased in the resulting signal.

The energy of the sidebands varies relatively to the energy of the carrier, but wait... this is the same than saying the DM is varying (without actually changing anything at the transmitter level).
The DDM between 150 and 90 Hz, is a function of the receiver offset from the centerline (in the range -5° to +5° the function is quasi-linear).

The space modulation actually creates a DDM
The alignment offset is determined by sensing the DDM in the resulting signal.
Antenna array
The LOC antenna is an array system, usually made of a large number of individual log periodic antennas, to form a narrow directional beam.

Localizer array at Melbourne airport, source
CSB and SBO are not distributed equally to all individual antennas of the array. Before reaching the antennas, their amplitude is tuned to form the three sub-beams, and the SBO signal is phase-shifted:

Individual amplifier-phase shifter-mixer
We have now all the basic elements to draw a LOC system:

Components of the ILS LOC system
Example with an array of two antennas (and final answer to the question)
Let's imagine two antennas fed with our earlier CSB and SBO signals with SBO phase being shifted by +90° for antenna 1 and by 270° for antenna 2.

Two antennas with phase shift of 90° and -90° (270°)
Let's see what happens when the receiver is on the centerline at P0, and at any point P1. For this simplified explanation to work, distance to LOC array (D) must be much larger than distance between antennas and centerline (d).
Receiver on the centerline
The received signal is the sum of the signals sent by the two antennas. As the distance traveled is the same, the phase at the receiver location is also the same.  The signal at point P0 on the centerline is:
$$S_{P0} = K_1 \cdot (CSB - SBO) + K_2 \cdot (CSB + SBO)$$
With $K_1$ and $K_2$ being the attenuation in free space, as the attenuation is the same for both antennas, this can be simplified:
$$S_{P0} = K_1 \cdot [(CSB - SBO) + (CSB + SBO)] = 2K_1 \cdot CSB$$
In CSB, the 90 Hz depth of modulation is the same than the 150 Hz one (see previous figure) There is no difference in depth of modulation (DDM) between the  tones: DDM = 0%.
Receiver at point P1
The distance traveled is shorter for antenna 1, and longer for antenna 2. A phase lead and a phase lag respectively appear for antenna 1 and antenna 2. The phase  lead is also equal to the phase lag. After some simplification that can be read in the linked thesis:
$$S_{P1} (\beta) = 2K_1 \cdot (\cos \beta \cdot CSB - \sin \beta \cdot SBO)$$
with $\beta$ being the angular distance from the centerline. This formula shows:

when $\beta$ is null (receiver on the centerline), we have the result already seen in the previous case,

when $\beta$ is negative or positive, the SBO signal influence increases and the overall depth of modulation between 150 and 90 Hz tones follows the same trend  until reaching -100% or +100%, which correspond to the absence of CSB signal influence at the receiver side.

Actual principle of the ILS, from this source.
The range of difference of depth modulation obtained must be mapped on the actual standardized values for LOC needle deviation, reducing the full range to a  standardized guidance angle. The LOC specification are found in ICAO annex 10 to the Chicago Convention, volume 1, attachment C:

This setting is obtained by tuning the amplifier of the phase-shifter from each individual antenna (the amplifier is actually common to all antennas). This amplifier  reduces the SBO signal influence on the DDM.
Clearance signal
So far we have assumed the localizer signal is only composed of the course guidance mixing the two original signals CSB and SBO. But it is very difficult to obtain  an actual highly directional beam, like the course beam, without creating unwanted lobes:

Radiation pattern of beam antenna, source
The problem with the secondary lobes is an aircraft approaching aligned with one of them will see it as being the main lobe, and will approach the runway using a  wrong course. There are some techniques to minimize these unwanted lobes, but none is practical for large arrays like a LOC system. Alternatives are to block the  unwanted waves after they are radiated, or to flood them, the solution used for ILS.
A replica of the course, the clearance signal, is sent with the course signal to outshine the secondary lobes.
The clearance signal has a power level of 10 dB less than the course level (a tenth of the power) and is radiated with a less directional pattern (less individual  antennas to form the beam):

Course and clearance
Today's localizers use a dual frequency system, this means the course and clearance signals are sent on separate frequencies.
When an ILS has a published VHF frequency f, the localizer course signal is actually sent on f + 4.75 kHz and the clearance on f - 4.75 kHz. The two signals can use  the same aerials or dedicated antennas in the array. Example of signals send to each of the 14 antennas of a Thales LLZ array (refer to above for what are course,  clearance, CSB and SBO signals):

Thales Dual Frequency – Localizer Antenna

Most of the information about ILS come from the 1983 thesis presented by Capt Dennis M. McCollum, B.S.. Respectful thanks to the author.

From the image it looks like the clearance signal cancels the side lobes but not the back lobe, which would make sense (because back-course approaches, etc.).  Also: Amazing answer.

@WayneConrad: Right. If the back lobe needed to be hidden, I believe other ways than flooding by clearance are available, e.g. reflectors on the antennas. Flooding would need more power in the clearance signal, as the back lobe is usually very energized, including for the log periodic antennas used.

That was clear enough that I sorta, kinda, more-or-less understood it! Is this the kind of detail that pilots are expected to know, or is this an area of interest/specialty of yours, @mins?

@FreeMan: Thanks a lot. This is more a topic of electronics engineering than pilot training. However, for a pilot it matters to know that if SBO is down, but CSB is live, the CDI will indicate "aligned" at any location, there was a serious incident based on this condition in July 2000.

Ok probably the best explanation of ILS going around! well done! I teach ILS to maintenance engineers and technicians,you could do my job-im sure!
Thanks for using my Space Modulation article from Wikipedia too!. although it's not completely wave interference/superposition, it's a lot more complex, but close enough.
There's DDM and SDM values once again were explained well-the Wikipedia articles for those are mine as well, LOL.
The DDM is determined by the combined resultant phasor (electronics version of vectors-used to represent signals rather than forces), addition/subtraction of the carrier wave power of the CSB and the two different frequency 90 and 150 hz modulation depths both 20% at the antenna and the SBO (sideband Only) power and 90 and 150 Hz modulation depths, again a phasor addition/subtraction !
When all those phases and powers combine in (air)space where the plane receives them, the ILS display effectively measures, a difference current in the indicator needle of between -150 microamps and +150 microamps.
if the plane is on centreline approach, the DDM is 20% 90Hz and 20% 150Hz = 0 % Difference between them, the needle remains centred-the plane is on centreline.
The Clearance is at a different frequency, but within the bandwidth of the avionics channel for the Course, (difficult concept) so it does not cancel the sidelobes-it swamps them! The Course sidelobes that are not within the navigable course path per-se.
The ILS concept is simple but the technology (from 1938!) is incredibly well engineered!
I've been teaching it for 15 years and EVERYtime I run a class, I learn something else about this system, I've worked on RADARs, MRI machines, CT scanners and Nuclear Medicine equipment-its up there in the technology stakes!

Welcome to Aviation.SE! Your answer starts like a comment referring to another answer. We prefer it here if answers are all standalone answers to the original question. Could you re-phrase this post accordingly? Once you have enough reputation, you can comment under the other answer...

Welcome Richard! Yes ILS and DVOR were really superbly designed, making the best use of reliable phase difference, at a time radio applications were mostly based on unreliable wave amplitude. Thanks for all your contributions on Wikipedia (I see I read many of the articles you contributed to), I hope you'll write here too.


text:
In what units is the Thrust-to-Weight ratio expressed?
is it correct to say the units for T/W is s^2/m?
This is what I did:
T/W= s/ft * ftlb/s * s^2/lbft

It doesn’t look right.  What units are you using for Thrust by itself and what units for weight by itself?

No. It should be unit-less. lb-f/lb-f = nothing

Yes, unitless.......

Thrust to weight would be a dimensionless ratio.  It sounds like you’re trying to write impulse to weight.

Newtons/Newtons = f/f = ma/ma ...

Thrust is a force. Weight is a force. Regardless of what unit one uses for each (Newtons, pounds, etc), the ratio of two forces is a pure number without any units. A T:W ratio of 1 means the thrust and the weight are equal, a ratio of 2 means that the thrust is double the weight, and so on.
It would be comprehensible to say "2 Newtons of Thrust per Newton of Weight", but "Newtons per Newton" isn't really useful, and that scenario would also be "2 pounds per pound," since the thrust is twice the weight. So simply "2" is how it's expressed - a pure number.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
follow up on the numbering system
with a runway 9/27 I am approaching from the east travelling in a westerly direction will I land at the 9 end or the 27 end?  in other words, is the number on the end the compass heading or the direction of travel?

Does this answer your question? How are runways numbered?

When I'm flying east, pointing east, with a compass heading of about 090 (i.e. 90 degrees), when I land straight ahead I will be landing on runway 9. Just after I pass over the runway threshold, at about 50 feet above the runway, I'll overfly the big "9" painted on the runway, and I'll touch down some hundreds of feet beyond it. If I then roll all the way to the far end, the eastern end, of the runway, I'll reach the "27" painted there.
Conversely, if I'm east of the runway & flying west, my compass heading will be about 270 and I'd land straight-in on runway 27. My compass heading as I'm landing (or taking off) corresponds to the designation of the runway that I'm using.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How do tailless aircraft yaw?
I was doing research on control surfaces present on an aircraft and I noticed that some aircraft like the B2 Spirit does not have a vertical tail stabilizer and therefore no rudder either.

How do aircraft such as these perform turns and counteract adverse yaw in banked curves? Do they apply some sort of differential thrust system for turning?

This answers you question: https://aviation.stackexchange.com/q/2252/42636

Technically the B-2 bomber has rudders, they are "drag rudders" (or split rudder) and are located on the outer portion of the wing:

Source: Quora
The rudders open in a clamshell configuration to create a drag force on the outer part of the wing. Because these are on the outer part of the wing, they can create a significant yaw force.


text:
How does the flight controls performance differ in a compressible and in compressible environment?
I know that in low speed flight compressibility effect is negligible and in high speed there will be an effect but how do they differ in flight controls such as primary and secondary?

The main issue with compressibility related to flight controls, is the shift afterwards of the Aerodynamic Centre.
Primary flight controls: At subsonic speeds the AC is at about 25% chord, which shifts to about 50% aft at supersonic speeds. The aftward pressure shift starts before the aeroplane can make the sound barrier: many early aeroplanes became uncontrollable at transsonic speeds and crashed.
Secondary flight controls: The B737 has the most secondary flight controls of the aircraft I've worked with:

Stab trim. The airplane trim shifts when entering transsonic flow, so the stab trim needs to be adjusted. Airplane certification makes sure that trim authority is sufficient even in a dive.
Rudder trim. Same as above, but usually yaw trim is not large during all-engine flight.
Throttle levers. I know of no effects of compressibility on the throttles.
Nose wheel tiller. Not used in compressible flow.
Trailing edge flaps. Not used in compressible flow.
Leading edge flaps. Not used in compressible flow.
Toebrakes. Not used in compressible flow.
Spoilers. Not used in compressible flow.
Landing gear levers. Not used in compressible flow.
Fuel shut-off levers. Not used in compressible flow.
Parking brake. Not used in compressible flow.

+1. Control reversal due to shock-induced separation could be added as a worst case scenario.


text:
How does a weak shock-wave boundary-layer interaction create wave drag (other than through direct shock losses)?
I'm trying to understand the physical cause of wave drag, beyond the simple statement "the presence of shockwaves increases the drag".
As far as I understand, in the case of a weak BLSWI (so without separation), the drag increase is due to:

Direct shock losses (momentum deficiency of freestream flow through shockwave)
A change in the state of the boundary layer due to the compression waves in side the BL

But how exactly does the second point work? If the BL becomes thicker after the SW, wouldn't that mean that the drag decreases because the velocity gradient and therefore the shear stress decrease?
I feel like I'm missing some tiny aspect...

Edit: I've found the following statement in this report, which maybe explains my problem better:

The
  increase
  in
  drag
  occurs
  directly
  because
  of
  the
  wave
  drag
  associated
  with
  the
  presence
  of
  shock
  waves.
  However,
  the
  drag
  also
  increases
  because
  the
  boundary
  layer
  thickness
  increases
  due
  to
  the
  sudden
  pressure
  rise
  on
  the
  surface
  due
  to
  the
  shock
  wave,
  which
  leads
  to
  increased
  profile
  drag.
  Lynch
  has
  estimated
  that
  at
  drag
  divergence
  the
  additional
  transonic
  drag
  is
  approximately
  evenly
  divided
  between
  the
  explicit
  shock
  drag
  and
  the
  shock
  induced
  additional
  profile
  drag

Why does the increase in BL thickness cause more drag? Shouldn't the velocity gradient at the wall reduce, leading in less shear stress? Or should I see it in a way that the already turbulent BL "takes up more energy in its vortical structures" from the flow, due to the increase in thickness?

I've seen a nice explanation of wave drag (and lift) around here showing that wave drag is simply because at supersonic speeds there is no pressure recovery beyond the thickest point. It is, therefore, not actually caused by shock waves themselves, just correlated with their presence, because where there is supersonic flow, there are shock waves.

So the effect you describe would be pressure drag, caused by a more negative pressure between the thickest point and the shock wave, when compared to the subsonic case? That would make sense. But still, the question remains, what's up with the thicker boundary layer? Shouldn't that even reduce the drag?

As you know, wall shear stress $\tau_w$ (and thus friction drag) can be described as:
$$ \tau_w = \mu \frac{\partial u}{\partial y} \vert_{y=0} $$
where $\mu$ is the dynamic viscosity of the fluid and $\frac{\partial u}{\partial y} \vert_{y=0}$ is the fluid velocity gradient at the wall.
The shock-wave boundary-layer interaction (SBLI) can change these two factors through several mechanisms:

A change in boundary-layer thickness: As you noted correctly, an increase in boundary-layer (BL) thickness will decrease the velocity gradient, compared to an otherwise similar BL velocity profile.
However, my conclusion after reading several publications is that the increase in BL thickness doesn't always occur. BL thickness seams to scale with shock strength and very weak shocks actually produce a thinner BL.
A decrease in fluid velocity: The shock-wave will obviously result in a decreased velocity outside of the boundary layer, which decreases the velocity gradient. This will also decrease wall shear stress, so it cannot be the dominant mechanism either, since it contradicts our experience of increased friction.
An increase in turbulence: The turbulence of the BL is significantly amplified by the shock-wave through different mechanisms.
The image below visualises the distribution of reynolds shear stresses, which are a measure for turbulence. A high magnitude (positive or negative) of reynolds stresses means high turbulence:

 Reynolds shear stress distribution of a SBLI. Notice the strong negative values in the lower BL behind the SBLI. Negative values indicate slow fluid moving upwards and fast fluid moving downwards. Also notice the value is zero very close to the wall, because the wall prevents vertical fluctuations. Image Source (I added the white markings)
The increased turbulence causes an increased momentum exchange between the lower and slower part of the BL with the higher and faster part. This momentum exchange changes the velocity profile of the BL. There are higher velocities closer to the wall. This results in a higher velocity gradient at the wall.
I believe this must be the main cause of increased wall shear stress. The stronger turbulence will probably decay downstream, so the velocity profile will slowly change back to the normal turbulent flat plate type.
Increase in pressure and temperature by the shock-wave: The rising temperature and rising pressure (not so much) increase the dynamic viscosity of the fluid and thus increase shear stress.

Most likely there are other factors that also play a role. I am no expert in SBLIs, I only read through a fair amount of literature to come to my conclusions.

Literature that I found especially helpful:

Anderson, J.: Fundamentals of Aerodynamics : My standard book for aerodynamics. Fairly easy to read. Section 9.10 briefly covers SBLI.
Schülein, E.: Skin-Friction and Heat Flux Measurements in Shock/Boundary-Layer Interaction Flows This paper shows the skin friction along an SBLI for different shock intensities. It provides schlieren images and figures that show the thickness of the boundary layer. It also discusses the causes for the observations.
Schülein, E.: Documentation of Two-Dimensional Impinging Shock/Turbulent Boundary Layer interaction Flow Here the velocity profile of the BL is measured at different locations for the same test setup as above. All the interesting data is presented in the appendix. You can see how the SBLI changes the velocity profile and BL thickness. It also explains how to calculate the skin friction from the velocity profile data (quite complicated)

Humble, R. Et al.: PIV Measurements of a Shock Wave/Turbulent Boundary Layer Interaction This is also the source of the image I used. They used Particle Image Velocimetry to measure the flow field of an SBLI. The part about turbulence is very interesting.
Anyiwo, J. and Bushnel, D.: Turbulence amplification in shock-wave boundary-layer interaction This paper investigates the mechanisms that amplify turbulences in an SBLI. Also very interesting.
Schlichting, H. and Gersten, K.: Boundary-Layer Theory Great book, every aspect of BLs. Detailed, long, harder to read than Anderson.

Great, thank you for the detailed answer! So, to summarise, the increase in drag due to a weak shockwave would be because of an increased turbulence in the BL after the shockwave (therefore friction drag)? I guess that the increase in pressure drag due to the lack of pressure recovery after the thickest point (see comment on my qeustion) also plays a role although that is only correlated to the presence of shock waves, and not caused by.

Exactly. The increased turbulence in the BL should be the main cause for increased friction drag. However, total-pressure loss across the shockwave always plays a role for drag. How much either effect contributes to shock-induced drag will greatly depend on the flow in each specific situation. Intuitively I would expect the total-pressure loss to be the stronger effect in most cases.

Alright, but the static pressure still increases across the shock wave, right? So intuitively I would think that this increased static pressure provides some "pressure thrust", if it acts after the thickest point.

Yes, static pressure increases across the shock wave. But in contrast to subsonic flow, supersonic flow accelerates when it expands.This is what happens after the thickest point. The static pressure decreases with flow acceleration. This is why you have an area of reduced pressure beyond the thickest point. Because of the total pressure loss, static pressure is even lower than how it would have been without the shock, if that makes sense.


text:
Can transition on an airfoil occur without prior separation?
i.e. can transition occur without the presence of a laminar separation bubble? If yes, what would this look like in a pressure distribution?

Here, a clear laminar separation bubble can be seen.

The Reynolds number has been increased by a factor of 10. The transition point is still visible, but it is not so clear anymore whether there is a separation bubble. Is there a possible scenario where a laminar boundary layer transitions without first separating?
EDIT: I'm not asking about how to force transition in XFoil or in real life (using turbulators), but rather if "pure" transition can occur naturally on an airfoil.

use a forced transition.

Good reference separation bubble. Reynolds Number indeed seems to be a critical factor, but laminar bubbles may be controlled with leading edge slats at higher AOA and/or with type of airfoil you use.  As far as controlling transition, as Peter said ...

And one for models by googling "laminar separation bubbles"  by MH-AeroTools.

I'm not asking about how to force transition in XFoil or in real life (using tuubulators), but rather if "pure" transition can occur naturally on an airfoil.

@Daniel what scale are you working on?  Full scale aircraft will be on the order of 2 million up to over 50 million.  Slow hand thrown gliders are around 20,000.

Any scale, I'm just wondering if this can occur at all.

Birds come in around 20,000 to over 100,000. And feathers act as natural turbulators.  As your graph suggests, look to higher Reynolds numbers and cambered surfaces.  It might help to narrow down your application, as AOA over a range will be an important parameter next.  The "bubble" seems to be a desirable area of lower pressure if it is relatively stable.  This may depend a lot on wing design, including sweep or washout.  Many good writers here speak of the dangers of the whole "bubble" "letting go" all at once.  This would be a sharp, dangerous stall.  A warning "buffet" helps.

So you're saying it doesn't happen on birds because of natural turbulators? Alright, what about a perfectly clean aircraft wing at low Re, can we have natural transition there?

Yes, Daniel, a bubble-free transition is completely normal. Only then the transition is forced by roughness instead of increasing Tollmien-Schlichting waves. Again, a forced transition.

Ok, that makes sense. What would it look like on a Cp graph?

Much better L/D with higher Reynolds number $1 × 10^6$!  That's why birds "force" transition.


text:
Who is responsible to check for radar-altimeter disagreement?
I have searched the site but failed to find a similar question. I know that modern civil aircraft are equipped with multiple radar altimeters. Question is: In the unlikely event of those outputting different values, whose responsibility is it to determine that the values are different? Is it the pilots, or the aircraft is required to warn? Does it depend on the aircraft type?
The question is mostly concerned about this accident: https://en.wikipedia.org/wiki/Turkish_Airlines_Flight_1951

That depends on regulations but broadly speaking it is ultimately always the Pilot In Commands responsibility to make sure all instruments are functioning properly.

§ 91.3 Responsibility and authority of the pilot in command.
(a) The
  pilot in command of an aircraft is directly responsible for, and is
  the final authority as to, the operation of that aircraft.

Now your probobly thinking "But if the autopilot is flying then who's really in command" well its the opinion of the FAA that even if the autopilot is taking care of everything you are still responsible for the aircraft and for making sure everything is working correctly. This question covers that more in depth.
As for things like warnings, alerts and auto-disconnects that depends on the specific system in question and the desired failure mode. For example if you have 3 radar altimeter inputs to an autopilot/auto-throttle the system can reasonably detect if one is erroneous and use the other two. There is lots out there on triple redundancy which is the basis of this.
What if the aircraft starts barking alarms at you saying "TOO LOW!, GEAR!" how can you even check that aside from cross checking your altimeters?

You can check your GPS altitude if available, while not a legal source for navigation you bet its a good backup.
If you are flying a stabilized approach (and in a plane like that you really should be) you should, at any given time know generally where you are on the approach and if you are on the glide slope (when available) you can gain a pretty good idea of your altitude.
In VMC, just look out the window!

I read the question and started to scroll down to give the unsatisfactory, but ultimate, answer of "the PIC", but you beat me to it. And with more detail than what I was thinking. :-)


text:
Why manufacturers does not producing variable sweep wing in our current scenario?
I read about many types of aircraft wing configurations like rectangular, swept forward, swept backword, and Delta wing. These configurations have their own advantages and disadvantages. But why we don't have variable sweep wing so that we can gain the advantages and better efficiency.

Because it's heavy and there are other better solutions. e.g. mid air refueling solves both the takeoff speed and mission range issue.

This was, as you are likely aware, the swing-wing was a major design element in the 1960s. Everything from the F-111 to the B-1 to the 2707 used them.
The design does have the advantages claimed for it, notably the better performance over a wider performance envelope. But the weight was a killer - not directly (although that may be the case in the 2707) but in relative terms.
Invariably what happened was that other parts of the system were improving too, engines, avionics, etc. At some point the performance improvement of the swing-wing required more weight than the same improvement from a new engine.
So it's not simply that it was too heavy, but too heavy compared to other moving goalposts.

Variable sweep is the best option when the top design speed is above Mach 2. If the speed requirement is reduced, a fixed, swept wing will be the better compromise, because it is lighter and less complex.
The quest for ever higher speeds made swing wings popular in the early Sixties, when engine performance, materials and engineering knowledge allowed to design aircraft which could operate at Mach 2+. Once the military analysts realised that this high speed was not needed, the swing wing quickly went out of fashion.


text:
What is the missed approach procedure when initiated below MDA/DA/DH?
Let's say you continue below the DA/DH on an IAP because you have the APL insight; and of course no lower then 100 ft above TDZE.
Now you arrive and don't see any required visual cues to continue to land.

When would you decide to go missed on an ILS? since you already descended below your DA
If you were to execute a missed procedure below the DA/DH on the same scenario above are you assured obstacle clearance? (my thoughts are no, obstacle clearance is only assured if its executed at the MAP, at or above MDA or DA/DH with a 200ft/nm climb)

When would you go missed if below MDA/DA(H)?
You're only allowed to operate below the MDA/DA(H) if all of the following are true (14 CFR 91.175(c)):

You can land using normal descent rate and maneuvers
You have at least the flight visibility required for the approach
You have at least one visual reference from the list (not just the approach lighting system)

91.175(e) says that you must go missed if those conditions are not met and you're below the MDA, at the MAP, at the DA/DH, or "at any time after that until touchdown", i.e. below MAP/DA(H). Essentially, if any of the conditions in 91.175(c) are no longer true, then you have to go missed immediately. Practically, the most likely issue is seeing the runway environment, but you should also go missed if your approach is unstable, for example.
On an ILS approach specifically, that would most likely mean that you saw the approach lights at DA/DH (or above), followed the glideslope down to 100' above TDZE, but then didn't see any more visual references. That should be very unlikely if you really do have the flight visibility required for the approach but I'm sure it could happen somehow, e.g. there's ground fog in the wrong place at the wrong time. Or something else could go wrong: the runway lighting completely fails because of a power cut, or there's an obstruction on the runway.
Is obstacle clearance assured if going missed below MDA/DA(H)?
No. The Instrument Procedures Handbook (chapter 4) says several times that if you go missed below MDA/DA(H) then you have no guarantees and it gives some general advice on what to do:

If the aircraft initiates a missed approach at a point other than the
  missed approach point, from below MDA or DA (H), or on a circling
  approach, obstacle clearance is not provided by following the
  published missed approach procedure, nor is separation assured from
  other air traffic in the vicinity.
[...]
In the event a balked (rejected) landing occurs at a position other
  than the published missed approach point, the pilot should contact ATC
  as soon as possible to obtain an amended clearance. If unable to
  contact ATC for any reason, the pilot should attempt to re−intercept a
  published segment of the missed approach and comply with route and
  altitude instructions. If unable to contact ATC, and in the pilot’s
  judgment it is no longer appropriate to fly the published missed
  approach procedure, then consider either maintaining visual conditions
  (if possible) and reattempt a landing, or a circle−climb over the
  airport.

This part is directed at pilots using night vision systems, but it looks like good advice for any pilot considering going below MDA/DA(H):

If a pilot initiates a go-around at a point below DA/DH or after the
  MAP, obstacle clearance is not necessarily provided by following the
  published missed approach procedure. Prior planning is recommended and
  should include contingencies between the published MAP and touchdown
  with reference to obstacle clearance, aircraft performance, and
  alternate escape plans.

I don’t really know the answer to 1), but you’re correct on 2) - obstacle clearance is not assured by regulations for a go-around below minima. However, due to all-engine aircraft performance most of the time far exceeding the procedure design assumptions, obstacle clearance is probably still fine operationally (at least in virtually all large commercial aircraft - can’t tell about small craft and GA), as long as they are flown properly and don’t lose an engine between descending through DA and re-gaining a height corresponding to the original obstacle clearance surface.


text:
How to find coordinates of the geometric center of a NACA 4-digit airfoil?
For a project that my team and I are working on, we are using Autodesk Inventor Professional 2018 (student license) for 3D modelling. We are working on NACA airfoils.
I need to create an airfoil within an airfoil so that the thickness distribution is uniform at around (10/12/15)mm.
To do this modelling, I've used two methods, neither of which is satisfactory. Please advise if using geometric center of airfoil is better than aerodynamic center. And how to find it. Please suggest any other method to get the uniform thickness distribution.

Use a NACA 6415, c=600mm and NACA 6412, c=580mm. Superimpose their aerodynamic centres.

The issue with this method is that thickness is more at Trailing Edge (TE) than at Leading Edge (LE).

Use a NACA 6415, c=600mm and scale it down to 0.85 with respect to the aerodynamic centre.

Issue with this is that the thickness variation is too high.

Cross posted at researchgate

Hello user27602, welcome to Aviation.SE. I see that at researchgate you have added images to clarify your problem. Can you add them here as well please?

I was trying to do the same thing last year but was using either the mathematical formulation of the NACA 4-digit profile or CATIA (for which, DS seems to be giving away a free academic license every October).  While I needed to do some special handling of the LE and TE areas, I could make a thick skin at the thickness I wanted.  It's not Inventor, but you could import the points and spline them from the mathematical formulation.  Would that be of interest?

The CAD software should have a function for this that's independent of the shape. This is often called "offset". e.g. for Inventor, https://www.youtube.com/watch?v=JaXqqy4o2mQ

I guess you wish to create a halo.
You can use Airfoil Tools' NACA 4 digit generator ( http://airfoiltools.com/airfoil/naca4digit ) to generate your NACA shape. It also features a plotter were you can set chord and halo thickness.
Positive values for internal halo and negative for external.
Internal halos are interesting to build wing ribbons.
Airfoil Tools generated NACA 6415
NACA 6415 600mm chord with 15mm halo (JPG)

Airfoil Tool's NACA 4 Digit generator

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are the missions of high-endurance aircraft?
What have been the missions of some notable high-endurance aircraft -- ones that can stay airborne for several days. Obviously, high-endurance missions have refueling, food, and human factors concerns that have to be addressed; what have been the missions that necessitated high-endurance aircraft?

Like chasing a submarine, monitoring airspace, etc. I think one of the most useful things it can do is to carry cruise missiles and fly "not-so-near" to another's coast line and just hangs out there for a while. Russia does this to Japan pretty often. You could argue long range BM would work, too, but cruise missiles has its own advantages over BM, and being able to launch up front enhances those significantly, and being able to stay ready to launch up front multiplies that "threat factor" even more.

Thank you for answering my question. Actually, I am writing an article on advantages of high endurance aircrafts. The problem point is why would anyone want to fly 500 (or more) hours nonstop when we are slowly inching towards hypersonic era when it may soon be possible to travel from London to New York in one hour.

Historically long duration flights were simply a way to fly further. Before the jet age crossing oceans was a time consuming affair and required what some may consider a "high endurance" aircraft/flight. Keep in mind it took Charles Linbergh in the Spirit of St. Louis 33.5 hours to cross the Atlantic and this was done with no refueling.

On the scientific/record setting front
A variety of airframes have been used for setting world records arguably a stunt in some cases and others with the pursuit of scientific progress. The current flight endurance record is held by  Robert Timm and John Cook at a whopping 64:22:19:05 (thats 64 days, 22 hours, 19 minutes and 5 seconds) in a Cessna 172 of all airplanes quite an endorsement for the endurance of the airframe/engine combo.
Solar Impulse a modern experimental solar aircraft that recently flew around the world regularly flew 15-22 hour legs due to its slow airspeed. In this case the mission effectively required long legs due to the aircrafts performance. This was a purpose built plane.
The Rutan Voyager was another purpose built aircraft that became the first airplane to circumnavigate the globe, non stop and with out refueling remaining aloft for 216 hours 03 minutes and 44 seconds.

On the operational front:
The Air Force operated Boeing VC-25 (often called "Air Force Once") is capable of aerial refueling and could remain aloft for quite some time and is built to be a mobile command center should the need arise for it.
The Boeing E-3 Sentry commonly known as AWACS provides aerial radar support and is capable of air to air refueling allowing it to stay aloft for well above its standard 8 hour endurance. I dont know a whole lot about its standard mission so Im not sure it ever flys multi-day missions.
There is a nice list of long range airliners here all of which have considerable endurance (but dont meet your multi-day requirement) and are all built (more or less) for servicing non stop routes from far locations.
Its worth noting that the general civilian use case of and airplane does not really lend to high endurance planes (multi day kinda stuff) in practice. Planes are generally for moving people and cargo in a timely manner and not really for taking off and just loitering for a while. The various militaries of the world simply move their aircraft around on aircraft carriers and launch them closer to where they are needed or fly high speed aircraft from farther away when needed. Airplanes also have the luxury (if permitted by laws and airspace regs) to move directly from point to point greatly reducing the distance traveled to get between locations over other methods of transit, again cutting down on travel time.

Some consider the space shuttle an aircraft and it was capable of multi day missions for the purpose of space exploration, satellite service and delivery and ISS transportation.

Maybe it is worth to add naval observation aircraft like the P-2 Neptune, the Breguet Atlantique or the Lockheed P-3 and unmanned observation platforms like the Global Hawk or even the Scaled Composites Centurion. They were designed for long endurance.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How do I choose the correct airfoil for my UAV project?
I am not able to decide which airfoil to choose for my UAV project as I am confused. I need to lift at least 10kg of extra payload excluding the weight of the fuselage. I can afford a wingspan of 3m with a cruising speed of at least 70kmph. Also I would like to know if there is any software or application to decide the airfoil for me?

You need to solve $L=1/2 C_L \rho V^2 A$. First determine the lift coefficient required for your airfoil based on required lift, speed and area, then select the airfoil with this $C_L$. You know you need a lift force of $10kg \times 9.81 ms^{-2} = 98 N$ for the payload.

You'll need the weight of the entire aircraft and fuel load to be determined with some sort of accuracy beforehand, not just your payload. Then you can find a $C_L$ range that would work. From there you can optimize the $C_D$ and $Re$ of the airfoil to suit your performance needs.

I found few great websites for choosing the right airfoil, so pick what shape of airfoil you need, to to this website... http://www.airfieldmodels.com/information_source/math_and_science_of_model_aircraft/rc_aircraft_design/plotting_airfoils/about_airfoils.htm - To choose the right airfoil from a large database, go here... airfoiltools.com

There are two methods of aircraft aerodynamic design. One method is through using known airfoils (for wing and tail surfaces) and deciding other geometric parameters iteratively while meeting certain criteria for performance, stability, control, aeroelasticity, manufacturibility, etc.
The other method is deciding roughly on the size and then attempting to come up with an airfoil shape, that (again) meets certain criteria. This second sort of design process is named inverse design. Where you try to get a desired Pressure distribution through computational iterations.
There are open source and free software as well as commercial options for both methods. Search terms: Airfoil design.
Any general RC aircraft design guideline and relevant spreadsheets will speed up the learning process. Search terms: RC aircraft design guide.
For detail information about the theory and practice; please refer to aircraft aerodynamic design textbooks.
Hope this answer sheds some light onto this wonderful and exciting topic. Feel free to shoot specific questions.


text:
Why do we refer to "power" for turboprop engines and "thrust" for turbojet/fan engines?
In jet engines, I read that thrust is related to the fuel flow rate, whereas in turboprop engines power produced is related to the fuel flow rate. What is the reason and brief math behind this?

Ultimately what you want from all three types of engines is quantification of thrust available to push an airplane through the sky. The turbofan/jet engines are self contained and produce thrust directly but a turboprop engine requires the addition of a propeller, which may have differing characteristics based on the installation.
Since thrust is not known for a turboprop until its installation has been determined, manufacturers instead quantify the power available to drive a prop. This allows engines to be compared so that an airframe manufacturer can make the proper selection.
Fuel flow is then related to either thrust or power as a measure of efficiency, depending on type of engine. There is generally no published mathematical relationship between power and thrust for a given engine. It is not needed for turbofans/jets and not possible to determine at time of manufacture for turboprops.

This is the only post that directly answers the stated question.

Jet engines directly produce thrust by exhausting gas (and in a modern turbo fan also moving a lot of air around them), so fuel flow rate is directly related to the thrust that is generated.
In a turbo prop the engine produces power which, via a gear box spins a propeller that generates the thrust. Since most turbo props have the ability to adjust the propeller pitch the engines power output is not always directly related to the thrust generated at a given time.

why you wrote that turboprop engines generate power, whereas while writing for jet engines you wrote that they generate thrust. This was my actual question?  I understand mechanics of both, but why is it intentionally written different? Am I missing something?

@user5349 Turboprop engines generate shaft power: P = 2piM*n via a low pressure turbine which connects to a shaft. The shaft drives a propeller through a gearbox.

To expand on what Dave said, the term Power is used for turboprops because the thrust produced by the prop is a function of horsepower applied to the prop, that is, torque @ RPM.
The gas generator of a turboprop - the jet engine part - has its output indicated as a percentage of maximum torque it can apply to the propeller gearbox, whereas a pure jet engine, who's push results from the mass airflow accelerating through the engine, has its output indicated by Engine Pressure Ratio, the difference between the air pressure going in vs the air pressure going out.
Turbofans are kind of in the middle of the two, being sort of a turboprop with a fixed pitch many bladed propeller. Because the fan is fixed pitch and has no constant speed regulating ability, you don't need to know the torque being applied to it and it's sufficient to go by just fan RPM (N1, indicated as percent of max).  Turbofans also show core engine RPM (N2), but the fan speed N1 is the primary power setting measurement.
For turbofans vs turboprops, its similar to how piston airplanes with fixed pitch props just measure RPM, like a turbofan, whereas piston planes with constant speed props need to show RPM and manifold pressure (MP being more or less equivalent to torque in a turboprop).
Not sure what you mean by the math part.

In turboprop engine, Is it that this torque produced by gas generator is constant ? the ultimate thrust generated by the power plant will be depending on the propeller pitch ? Which is a variable ?

No the torque that acts on the propeller varies with gas generator speed.  Think of a turboprop as a pure jet engine where you point the tail pipe at a windmill to make it spin.  You could call the jet engine part a "gas generator".  The windmill getting blown on is connected to a huge fan by a gearbox.  The air blowing on the windmill generates torque to make it spin, driving the fan.  You speed up the jet engine itself to increase the torque on the power turbine (the windmill). Move the windmill to inside the jet engine to just behind the compressor's turbine, and there's your tubroprop.

@John K Your opinion please.  Thrust produced by the prop (propulsion force) is determined by pitch and rpm of prop.  Power to turn prop = DRAG of prop.  Would not Power to turn prop be THRUST of pistons no matter how it was geared?  Would not the Propulsion force actually be the LIFT of the propeller.  Just being academic, but it is interesting to consider in light of the trend towards efficiency in jets (fans are getting bigger and slower).

I would say that difference is due to the fact that the thrust of the jet pretty much all is from the air that is forced to accelerate out the convergent tailpipe nozzle, whereas the thrust of the propeller is a secondary result of torque applied the propeller.  So, the go lever in a jet just controls how fast the air comes out, whereas the go lever in a TP controls the torque to the gearbox, and torque x RPM = HP, so POWER is a more appropriate term.  With turbofans you have a kind of half TP and half jet, but there is no measurement of torque vs RPM for the fan so thrust lever does the job.

The real reason is that, by definition, a jet engine does not output any kind of power but outputs the thrust force instead.   Turbopropeller engines do not output a thrust force but rather output a mechanical power through a shaft and gearbox supplied by said power by either a power turbine or the gas core itself.
That arrangement has to do more with how the FAA and other governing bodies defined these engines, not so much based on physics.   I am aware that both engines are, in fact, generating energy and power, it’s just we don’t typically quantify energy output from jet engines that way.
Ironically many turbopropeller engines are often rated in equivalent shaft horsepower (ESHP) this is a sum of both the mechanical power delivered by the driveshaft and the reaction thrust created by the engines exhaust gases.
This also has regulatory impacts as well.  Pilots of jet aircraft do not require neither a complex nor high performance endorsement in their logbooks to log PIC time in a jet, but it is required if your aircraft is powered by either a reciprocating or turbopropeller power plant.

Agree 100%.  This appears to be a matter of convention, a little like saying, ok, from now on we're going metric.  Nope, not happening.  Maybe, in the future...

Definitely not ironic for turboprops to rated ESHP as mechanical power + engine exhaust thrust.  This is in fact correct.  Shaft horsepower is torque at a given rpm, which for piston engines is the AVERAGE thrust (cylinders firing per minute x Lever Arm of crankshaft divided by diameter of drive shaft x any other gearing ratio).  At constant rpm this = DRAG of prop (thrust = drag) and produces a forward force WRT aircraft.  Thanks for allowing my opinions to be aired.  I will try to learn the conventions.

Oau, a lot of answers and almost nobody touching on the real issue.
Propellers will produce a thrust that decreases with increased speed. Since power= thrust x speed, this means the engines will produce (more or less) an (almost) constant power throughout the speed range of the aircraft. Given a specific fuel flow, it will produce a specific (calculatable) power, which will translate to a varying thrust depending on the current speed. Therefore fuel gives you power.
Jet engines will produce a constant thrust (almost) regardless of current aircraft speed. This means “power” is not really a usefull measurement in jet (all of them are thrust-rated, not power-rated), as you can just double the current speed and the jet is now producing twice the power, even though you have not changed the fuel flow. A specific fuel flow to the jet engine will produce a specific (calculatable) thrust, but it can produce any amount of power depending on the current velocity of the aircraft. Therefore fuel gives you thrust

You wrote in a comment on Dave's answer:

why you wrote that turboprop engines generate power, whereas while writing for jet engines you wrote that they generate thrust. This was my actual question? I understand mechanics of both, but why is it intentionally written different? Am I missing something?

So it sounds like the main question you're trying to ask is: what is the difference between power and thrust?
Power and thrust are two different measures of the output of an engine (just like how height and weight are two different measures of the size of a person). All engines generate power, and all engines generate thrust, but they're two different numbers with two different meanings.
Thrust is a bit easier to describe. Thrust is often measured in pounds or newtons. If an engine is producing 500 pounds of thrust, then it's pushing on the aircraft with the same amount of force that a 500-pound weight would push on it. The difference is that the engine pushes forwards while the weight pushes downwards.
Output power is a bit harder to describe. Power is often measured in horsepower or kilowatts.
For rocket engines, the formula for output power is simple:
$$\text{output power} = \text{thrust} \times \text{exhaust speed}.$$
Unfortunately, most airplanes aren't powered by rocket engines, so this simple formula no longer applies, though. But the basic principles are the same:

As an engine produces more thrust, it will also produce more power.
Even if an engine is producing a constant amount of thrust, as the airplane flies faster, the engine will end up producing more power.

Power production is directly related to power consumption. If you make an engine produce more power, then it will consume more fuel.

I see that this answer has gotten a downvote. As far as I know, there's nothing wrong with it; it seems correct and useful.

I was not the downvote, but it might be because of the ‘power= thrust x exhaust’. For any thrust, , force x velocity (of that force)  will produce power

@Radu094 Does that contradict what I said? If it does, I don't see how.

Not contradicting, just unclear I’d guess. A sitting (fixed/bolted) rocket spweing exhaust at high speed produces a nice display and a lot of thrust, but 0 power. Not really related to the asked question, I know, but it caught my eye and someone prob downvoted because of it

@Radu094 That's a good point. I want to say that the rocket still is producing a lot of power, but all of it is going into the exhaust. But I don't know how the "output power" of an engine is usually defined.

Glad you folks are seeing it.  How far it moves has nothing to do with thrust or power.  Definition is by convention, and it was good to go back to James Watt to see what was going on. P=thrust/time.  How many cylinders fire per minute.  How much fuel the jet uses per minute.  Interestingly, a jet WILL have more power with speed due to air ram effects, exactly the same as a supercharged piston engine!  More fuel!  Will not pursue any further, enjoyed sharing it with you all.

It is because in a turboprop, almost all of the energy extracted by the engine turbines is used to run a propeller. So, it is all about how much power that is required to turn the propeller. To measure the amount of force required to rotate the propeller, we use a sensor to measure the amount of twisting force acting on the propeller shaft. This sensor is called a torque sensor and in almost all turboprops, torque is the indication used by the pilots to control the engine power. But we all know that according to the power equation:
Power = Torque x RPM
From the equation we understand that power can be increased by either increasing the torque or the RPM. So, how do turboprop pilots only use torque to control the engine power? Well, it is simple. In a typical turboprop engine, the propellers are variable pitch or you could say that they are run at a constant speed or RPM. This is achieved by a propeller control unit (PCU). In simple terms, as the pilot pushes the throttles forwards, and as the torque increases, the RPM remains constant at the pilot set value using the condition levers. The way it is done is very simple. As the pilot increases power or torque, the propeller tries to run at a higher RPM. But the PCU senses this and increases the propeller blade angle and slows it down until the set RPM is reached. Similarly, if you try to control the propeller RPM using the condition levers there will be a change in the torque. That is, if you pull back the condition levers (increase blade angle) to reduce propeller RPM, there will be a corresponding increase in engine torque. So, at high torque settings, you have to be very careful because you could over torque the engine. When we set power in turboprops and if requires a reduction in propeller RPM, we reduce the torque using the power levers first by a bit before reducing RPM using the condition levers. This is particularly important, when setting climb power after a take off which is done close to the maximum allowable engine torque.
So, in a turbprop engine, the combination of torque and propeller RPM gives a certain amount of shaft horse power (SHP). At full RPM and at full torque setting the engine will produce the maximum SHP. In a typical take off, the engine produces an SHP that is very close to this maximum SHP.

On the other hand, the turbofan engines measure pure thrust output. Some engines measure it as a percentage of low pressure compressor RPM (N1 percent). Why do we use N1? Because low pressure compressor or the fan produces most of the thrust in a turbofan. The amount of thrust produced by the hot section is quite small compared to the thrust produced by the fan or the bypass. In some other turbofans including old turbojets, the thrust measurement is given as a pressure ratio or more specifically an Engine pressure ratio (EPR). This is a more direct measure of engine thrust, as it is the ratio of engine exit pressure over inlet pressure.

I thought even the engine design would be a bit different depending on your application.  For a pure turbojet, you want maximum thrust out the rear, so maximizing torque turning a shaft isn't the goal.    On the other side of the spectrum, going past the turbo fan to the turbo prop or even max torque application like a turbine generator, you want to minimize thrust and convert all expansion gas energy to torque.
That is why most turboprops use centrifugal compressors at at least one stage, and pure turbo jet engines use axial compressors.   The former converts most of the expanding gas energy to circular motion (we will call it torque) and the latter accelerates that expanding gas for maximum air/gas velocity out the back for thrust.
A turbocharger in an auto engine or the turbo charger in a piston aircraft engine use centrifugal compressors similarly because they want to generate circular motion (torque) on an air compressor to increase the amount of oxygen in the combustion chamber.
I'm not sure if this is correct, because I just thought of it, but I did some research to confirm that pure jets want thrust, and axial flow, and high gas velocity at exhaust, and turboprops want torque, to turn a shaft into a psru then a propeller, so these engines have different compressors that convert the energy into different types of motion.

Posts made as 'Answers' should provide an answer, not a guess.


text:
Is there a limit for which vision can be corrected to 20/20 vision for being a pilot?
I have 20/400 vision and can be corrected to 20/20 vision with glasses.
Does having 20/400 vision disqualify me from getting a pilot license?
Would it prevent me from getting the required medical certificate?

*prevent me from getting a medical certificate?

Which country or regulations are you asking about?

India. If you don't know the regulations for India, tell me for any country.

You can find the answer for the US/FAA regulations here

Related


text:
Where can I do an FAA IPC in Canada?
I have an FAA Commercial License and my instrument checkride was completed 8 months ago. Now I am in Canada and trying to convert my license. Problem is the currency of the instrument rating. Transport Canada said they require an IPC check and I want to know if there are any FAA CFIIs here in Canada who I can do the IPC with. I don't want to go back to the US just to do that.
I was reading the other similar post by a user who says he signed an IPC for FAA pilots in Canada. If you are that user and are reading this post please let me know where I can do my FAA IPC in Canada.

The OP quotes the suggested-as-duplicate post, so he's pretty clearly aware of it, and it didn't answer his question. Don't see this as a dupe.

Note that @mongo said he or she signs off approaches, nothing about IPCs. To complete an IPC you will need a CFII.

This isn't an answer and you may already have looked into it, but... The Canadian requirement is to have done an IPC in the previous 24 months, but your checkride was only 8 months ago. Have you asked them if they'll accept the checkride in lieu of the IPC?

I know the requirements and been searching the FAR/AIM as well and the problem is, the way it is worded in  the FAA FAR/AIM about the IFR currency and Transport Canada is only going by what is written over there which misguided almost every single pilot. When i told transport canada that i dont need IPC until 12 months from my flight test according to FAA, they are just not convinced. There is a grace period of 6 months after your first 6 months of currency period, now the grace period is not mentioned in simple language, thats the problem.

I get this  question a lot.
I am an FAA airplane and helicopter CFI with an Airplane "double eye" the Instrument Instructor and Multi-Engine instructor. In Canada I'm also an airplane ATPL and airplane  flight instructor.
You can see the certificates for any FAA Instructor with an Instrument Instructor rating at the FAA Airman Registry https://amsrvs.registry.faa.gov/airmeninquiry/
Just enter first name ,  last name, click on  the name and you will see all the FAA Instructor certificates. Unfortunately, Transport Canada does not have such a public pilot data base.
Under an agreement between TC and  the FAA, a pilot who holds BOTH an FAA pilot certificate with an instrument rating, and, a Transport Canada pilot  licence with an instrument rating can use the FAA IPC to use their Canadian Instrument Rating.
There are some differences.
(1) The FAA IPC cannot be done entirely in a simulator, all the approaches with the exception of  the circling approach can be done in a simulator.
(2) A Canadian IPC can be done entirely in  an approved simulator.
An FAA instrument instructor is authorized to do Instrument Proficiency Checks for FAA pilots.
If you hold only a Canadian Pilot Licence without both an FAA Pilot Certificate and TC certificate, you can only do your IPC with a TC authorized person.
Here is where it gets interesting.
Whats required for a Canadian Pilot to get an  FAA Pilot Certificate and Instrument Rating.
Not much.
You  need to pass the TC to FAA pilot certificate exam which is probably 25 questions. You also need to pass the TC to FAA Instrument written exam.
You need  a medical and an  application completed.
You also need your identity confirmed by at least an FAA Instructor. Any FAA Instructor can check  a person's identity for the purpose of obtaining a student pilot licence or higher rating.
Its easy enough to do a cross border trip to get your FAA medical  or  you can choose one of the few that exist in Canada. You will need to do the written exams in the USA as there does not exist any FAA examination centre in Canada.
In sum, it can be done very quickly.
In one day, you can do the FAA medical, do  the exams, fill in an application and get your identity confirmed.

You will need to find a US CFI-AI who can perform a competency check for you.  It can be done in the US or in Canada (in a US registered aircraft unless that CFI holds TC and FAA credentials).
If you are within 6 months of being current, you can do approaches with a safety pilot in Canada. If you are flying a US registered plane, then you will most likely need a US certificated safety pilot.  If you are flying a Canadian plane, then your safety pilot should be a Canadian pilot, acting as PIC for the flight.  You could still log the approaches and simulated instrument time.
In my experience, it will be far easier to find a US CFI in Canada than it is to find a Canadian CFI in the US.  Ask around.  Consider contacting an organization like AOPA if you are a member, or COPA, and asking them for help locating US CFIs.  I would also ask if they have additional guidance because they may have letters from Transport Canada and the FAA which may have policy considerations which might be good to know.

One additional thought...at flight safety, there were IPs who had multiple country certificates.  If you are near a training hub, you might look into that. It will probably be expensive, but flying a Level D simulator is lots of fun. Practically, however, you should look for a cheap trip to the states, get your IPC done. While you are at it get a flight review endorsement, so you aren't stuck trying to solve that problem later.

I talked to an CFII from my school and he said I don't have to do the IPC because under 1 year of the last instrument flight test i don't have to go for IPC instead do the regular currency requirement with safety pilot or in the simulator and that also i can do it in CANADA with any school. The problem is the wording in FAR/AIM which doesn't say this grace period. Everyone there is being taught like the way i was told and even in the flight test the examiner ask you this recency question and the answer is IPC after 12 months of initial flight test. PILOTCAFE is the material used for oral exam.

The issue is whether TC wants you CURRENT or not.  If they want you current, you can still accomplish the approaches, etc. Since you don't have a Canadian certificate you will need a Canadian safety pilot, who will act as PIC.

For the conversion process TC don't, according to https://tc.canada.ca/en/aviation/reference-centre/advisory-circulars/advisory-circular-ac-no-401-001 §6.4. They require IPC in the past 24 months prior to application.


text:
Should the route be changed if the GPS RAIM prediction fails?
Do i have to change the route if the GPS RAIM prediction has one or more outages along my route?

The answer might be affected by the route you filed and your equipment... flying Victor airways with a VOR receiver, probably not; Flying in RNP-3 airspace with INS units that can update off VOR/DME as well as GPS, probably not. Flying a Q route with only a GPS, maybe (more interesting question, at least). What sort of route + equipment do you have in mind?

@RalphJ Thanks Ralph. This question popped up when I was exploring aviation topics.I think your comment answers my question.

The answer to this question will probably depend on the equipment involved and type of route being flown; as asked, the question is open ended, and there are several types of operations where the answer is a definite "no" (i.e. you do not need to adjust your route), while in other situations, that answer might change.
If you are flying on Victor airways using VOR receivers, then the answer is certainly "no".  You aren't navigating with GPS, so you don't care about the RAIM.
If you are flying in RNP-3 or RNP-10 airspace (i.e. the required navigation performance is 3 NM or 10 NM -- so your INS can drift that far before you need to do something different) with an INS or multiple INS's, then you probably don't care about the RAIM... your INS's won't drift that far in several hours, and the RAIM should be back to acceptable limits well before then. (Maybe there are exceptions to this; I'm not aware of any.)
If you are flying an approach where GPS is your primary navigation, then YES, you absolutely do need to change your plans if the RAIM fails.  Break off that approach, go somewhere where the RAIM is good, use a conventional navaid for the approach, whatever the best option is for you.
I'm not aware of any enroute operations where RAIM is a factor; what I'm flying has dual INS's and the FMC can update its position off of not only the GPS but also from conventional (DME/DME, or VOR/DME) navaids, and we don't worry about the RAIM enroute, even when flying Q routes where GPS is required.  (Maybe that's a hole in our OpSpec, I dunno.)  We do look at it when flying an RNAV approach.

AIM 1-1-15, 5(h)(3):
If a RAIM failure/status annunciation occurs prior to the final approach waypoint, the approach shall not be completed since GPS no longer provides the required integrity.
Other text in this section indicates that the GPS receiver should be considered advisory only and directs you to the receiver operating manual, which I suspect will tell you to consider no RAIM = no reliable GPS.

Thanks Matt for the detailed answer.. is there any documentation link that i can go and read about all the information about RAIM?


text:
What are "ground-based navigation systems which provide vertical guidance" and how do they compare to altimeters?
I am a total layman when it comes to aviation but for a school presentation I am trying to understand the reasons why Air Canada 624 crashed while trying to land on runway 05 at Halifax (CYHZ). I have read the investigation report and the press release the Air Canada pilots union have published. The press release says:

ACPA strongly advocates for all major airports to have the necessary
  equipment to offer both lateral and vertical guidance and high
  intensity lighting to allow pilots to orient aircraft for landing.
Many runways at Canada’s largest airports lack the ground-based
  navigation systems which provide vertical guidance – that is, the
  distance between the aircraft and the ground – to orient an aircraft
  for landing.

I was under the impression that an altitude reading is always available through the radio altimeter or the barometric altimeter.
Why would such a "ground-based navigation system" that provides "distance between the aircraft and the ground" be needed? What can it provide that the regular altimeters on board cannot provide?

If the altimeter was used to approach the runway, the required altitude would need to be updated as the distance to the runway changes, this can only be done by a computer. The advantage of the ground guidance is that it doesn't need any calculation, or even to know the distance to the runway. It can be a light beam or a radio beam transmitted from the ground and the aircraft just follows it.

Thank you for your answer. I did not know that an ILS sends a beam that the aircraft can follow, this is new information to me.
About the need to calculate the required altitude as a function of distance, can't this be calculated beforehand and tabulated? It would solve the problem without needing expensive ground guidance systems, wouldn't it?

It could be pre-calculated. This would still be an additional workload and a distance measuring equipment would be required (on the ground and onboard). The information could also be displayed on the altimeter, but I believe at the time, displaying this additional information was too difficult. Ref the ILS beams (loc and glide), each beam is actually a pair of beams and the receiver in the aircraft senses the axis of symmetry, where the beams are received with an equal intensity.

They mean an ILS - Instrument Landing System.
An ILS consist of:

A localizer component, which tells the angle between the aircraft's position and the extended runway centerline. In short, it tells the pilot whether they are left or right.
A glideslope component, which tells the angle between the aircraft's position and the extended runway surface on the ground. Typical glide slope angle is 3 degrees. In short, it tells the pilot whether they are high or low.

An ILS is usually coupled with an DME (Distance Measure Equipment), which tells the distance between the aircraft and the runway threshold. Using the glideslope angle together with the horizontal distance, one can compute the height above runway level.

From page 12 of the report:

AC624 was then cleared for the straight-in LOC approach ...

Which means only the lateral (left or right) component is available, there is no signal which tells the pilot whether they are high or low relative to the ideal descent path.

A radio altimeter is not very useful when the approach path leading to the runway has uneven terrain. It tells the height above the ground directly under the aircraft.
A barometric altimeter needs to be calibrated by a reading at ground level. It tells the height above sea level.
Both of these altimeters cannot tell the height above runway level. Therefore the pilots must rely on other data to monitor the descent profile, which increases the pilots' workload. In a situation such as low visibility and high crosswinds, this presents a very challenging situation.
If both vertical and horizontal guidance is available, then the pilots can simply follow this guidance all the way to the runway. They can also program the autopilot to follow this guidance, if suitable equipment is installed on the ground as well as onboard the aircraft.

Note the green bars: one is horizontal and one is vertical. They form a cross on the display. If the horizontal bar is up a little bit, the pilot would pull up a bit. If the vertical bar is to the right, the pilot would turn right a bit. These green bars only possible when ILS is used.
It is possible to compute the relative height by using a DME and a barometric altimeter only. For example, I am now 5.5 nautical miles from the runway threshold, and according to the approach chart, I should be at 1,800 feet. Right now my altimeter reads 1,600 feet. That means I am too low and I should reduce my vertical speed to slow down my descent. This calculation has to be done by the pilot himself and not by a computer. Furthermore the pilot must fly the plane, i.e. adjust the control surfaces, run landing checklists, visually verify the alignment with the runway, talk to ATC, set the flaps etc. A pilot is trained to do all this, but increased workload also increases the chance of making an error.

Thank you for your answer. What if you know the height of the runway itself with respect to sea level? Couldn't you use the barometric altimeter reading by subtracting that height from the reading?

@TylerLungrin Yes, you very much can. Indeed, in the UK we actually have another Pressure Setting called the QFE which is used to show you the height above the airfield to save the maths. Most countries just use the local QNH, which is altitude (Above Sea Level) and the pilots can calculate as you say. This isn't accurate enough for Instrument Approaches however

@TylerLungrin yes, but this reading is not accurate enough for computers to calculate a vertical guidance. New aircrafts nowadays have the ability to use GPS height when ILS is not available. This trend is spreading, but it is not the majority yet.


text:
KC-130: What are the four tanks/tube on the wings in between the engines used for?
On the KC-130, what are the 4 tanks/tube on the wings in between the engines used for?
I've seen them in photos but without explanation. I cannot find any mention of them whatsoever. If not storage tanks then what else would they be used for?

The tanks between the engines on each wing are external fuel tanks. The pods outboard of the #1 and #4 engines are aerial refueling pods which house the hoses and drogues that are reeled out when refueling other aircraft.
This page has several examples of the outboard pods trailing drogues for refueling.  These pods are unique to KC- and MC-130 variants. One example:

The inboard external tanks (between engines 1 & 2 and 3 & 4) are found on most variants of the C-130.


text:
Battle Damage Checks / Procedures
Studying the damage model of DCS, especially on Warbirds, I encountered an operational doubt about the Battle Damage Checks.
Here is the question:
Suppose that after a sortie you are RTB, and you don't have the possibility to have a visual check coming from the ground and you don't have a wingman; your aircraft is particularly damaged and it is night.

Excluding the Flight Control check (that's pretty straight forward), what would you do about landing gear and flaps?
If you select the landing gear down there is the risk that it would not come back up, at least increasing drag and fuel consumption and the risk of injury in case of forced landing (if particularly damaged).
For the flaps, If you select them down, during the normal pre landing procedure, if one flap goes down and the other stays up, trying to perform a recovery or bail out would be very hard due the low altitude, but anyways it would put the aircraft in an undesired state.
If you don't move anything, you will perform a gear up, flaps up landing, the aircraft will be more damaged but probably repaired, and in theory, you will not expose yourself to other risks.

I know that is very type related, and it depends a lot of the situation, but what would you do? Do you have any additional information about the real checks and procedures that are done (or were) in these situations?

As requested in the comments, to facilitate answers and discussions, let's focus on the Spitfire MK IX.

@ZeissIkon I don't think it's that type-dependent actually. All those planes had gear down indicator and, alternate gear extension mechanism (even if different in each), and radio for asking the tower if your gear looks down (it it wasn't shot, of course; you'd always fly a pattern, because none of these planes had anything for an instrument apprach).

OP specifically mentioned night, no wingman, etc. -- question is for a pilot on his own without external help.

@ZeissIkon there is still the indication, the alternate extension, and the parachute on your back—you can bail out from 800 or 1000 ft you have on downwind.

@JanHudec The problem is that the indication that you have in cockpit is only about if the landing gear is down and locked, not about his condition, you can have the strut in the correct position and locked but missing a huge part, you will not know it until touchdown. Bail out is always the last option, you will not jump if the aircraft still flyable.

Excellent discussion topic, but any answers will be speculative and opinion based of a “here’s what I would do if…” nature.  It boils down to knowing your airplane, and making sound decisions based on risk assessment and good judgement.  And there’s no answer specific to the Spitfire either unless you can describe a single scenario where a specific system or condition might be in play.  (I.e. this question is both too specific, and too general…)

Note that you wouldn't find yourself in a battle-damaged Spitfire Mk.IV at night. In a Beaufighter, Mosquito, Wellington Halifax (all are multi-engine and multi-crew) yes, but not a Spitfire.

Regarding votes to close... while there may be no single checklist-style decision tree leading to "the answer," I think the comments show that it's entirely possible to provide a reasonably objective list of factors that a pilot in this situation would use to evaluate his decision & build his plan. How much the pilot might weigh this or that factor may reflect the pilot's opinions, but that doesn't make the answer here "opinion based". Voting to leave open.

@RalphJ I want also to add that, what would you do is just a small part of the question, if someone knows, (due experience or other sources), standard operating procedures to follow after combat is more than welcome in the answers.

Please explain what are "DCS" and "RTB"

@quietflyer DCS: Digital Combat Simulator, RTB: Return to base.

What can be answered for sure is that from an airmanship perspective, you will first of all get yourself near the airport at a suitable bail out altitude, say 2-3000 ft, where glide ratio is not a factor for something like a gear-stuck-down situation, before you try anything with the airplane's configuration.
Then you will assess the situation after trying out the systems, and make a decision whether to try to land or bail out, which will have to be judged on a case-by-case basis. A lot of pilots will be so terrified of jumping they will try to land the plane in just about any condition unless they have no choice.  Others will prefer to bail out.

I can offer some documented anecdotes. The Big Show has a chapter about accidents.
One accident it describes matches this question. A damaged Tempest was returning from a combat sortie. The gear didn't go down, one leg got stuck partially extended. The pilot called tower, they told him one leg is partially down, he tried to get it back in, tried alternate extension, nothing helped. Unfortunately he turned the radio off before they said he should bail out. The emergency landing didn't go well, the aircraft bounced, flipped over and started burning and the pilot died in the fire.
A bit later Pierre Clostermann (the author) got in similar situation. Gear wouldn't go down, and additionally his canopy got stuck (rails bent from bullet hits), so he couldn't bail out. I think his gear remained in, but in the Tempest the biggest risk came from the big cooling radiator under the nose. Fortunately that landing went better, the aircraft dug a deep furrow, but remained in one piece and he safely got out.
It also mentions other accidents caused by mistakes rather than damage, like when a fighter collided on final approach with a transport plane bringing some new recruits that was flying below. The fighter pilot does not see below well, was coming from behind where the transport pilot can't see, and the tower didn't notice in time to shoot a red flare to wave them off (landing clearance was usually given by green flare or denied with a red one).
Overall I think around half of the casualties was in accidents rather than directly in combat, with some of the accidents being caused by battle damage and some by mistakes. Not unexpected given many of the fighter pilots didn't have all that much experience. Often just basic flight training and something like seven hours on the actual fighter type before they got assigned to combat units.


text:
What is the reason for setting flaps 1 on the ground at high temperatures?
Yesterday I've seen an A320 standing at the gate with flaps 1. While this question answers, why the flaps should be up, is there any reason to retract the flaps to 1 instead of up?
In addition it seems to apply only in high temperatures (30°C, and rising yesterday). As well in this pilot eyes video the FO confirms the temperature (31°C) before setting flaps to 1 as they taxi to the gate.
I've seen this only on the A320 (yesterday) and on the A330 (in the video). Does this apply only on Airbus aircraft or as well on others?

Never heard of anything like that in the 737. I'm entirely mystified as to what that might accomplish.

Are you positive that you saw the flaps, and not the slats? (Compare ymb1's answer, which discusses the slats.)

I’ve been flying several types of Airbuses for the past decade and have never seen this kind of procedure before. There certainly is no such limitation in the manuals.
In general, it sometimes happen that there has been such a practice in a previous aircraft type the company has operated and these kind of procedures carry on as best practices or company procedures without any “real” reason.

Well, I stand corrected! See correct answer above.

Thank you, extrapolating procedures from an aircraft model to another does not comply with the rules of the art.


text:
How is Turbojet defined by the FAA?
During my commercial ground school today, we were asked if turbojet and turboprop aircraft are defined differently by the FAA. While I know the answer is yes, I don't know where in Title 14 the answer can be found. I've searched multiple resources for over two hours.
Thanks!

Welcome to Av.SE!

The FAA does not define the terms turboprop and turbojet in the FARs.  These terms are defined in the Pilots Handbook of Aeronautical Knowledge.

Turbine engine. An aircraft engine which consists of an air compressor, a combustion section, and a turbine. Thrust is produced by increasing the velocity of the air flowing
through the engine.

Turbocharger. An air compressor driven by exhaust gases, which increases the pressure of the air going into the engine through the carburetor or fuel injection system.

Turbofan engine. A fanlike turbojet engine designed to create additional thrust by diverting a secondary airflow around the combustion chamber.

Turbojet engine. A turbine engine which produces its thrust entirely by accelerating the air through the engine.

Turboprop engine. A turbine engine which drives a propeller through a reduction gearing arrangement. Most of the energy in the exhaust gases is converted into torque, rather than using its acceleration to drive the aircraft.


text:
Are there any civil programs that uses S1000D?
For civil programs, aircraft documentation (especially maintenance) used to follow ATA 100 or ATA iSpec 2200.
S1000D, on the other hand, is mostly used in military programs, but civil program may use this specification and clearly should supersede ATA-like documentation.
I found this document that talks about S1000D implementation in civil aviation, but it is quite old.
For now, do you know of any civil program (in production or currently in development) that effectively uses S1000D? Which version do they use?
To my knowledge, the only civil programs currently writing documentation in S1000D are Airbus A350 and Boeing 787. Are there any other ones?

According to https://www.aecinc.com/company/blog/2015/09/18/considering-s1000d-for-your-aircraft-technical-documentation, Bombardier C-Series uses S1000D as well


text:
When sizing a powerplant for a GA aircraft, how do you estimate the power required for other electrical systems?
When designing a GA aircraft, I have sized a powerplant (in kW) needed to run the plane on its mission. However, this does not consider the power needed for avionics, de-icing, etc. How do I approach finding the numbers for these systems? Are they significant?
For de-icing, this study from Hamburg University shows some example power consumptions from 23 KW/m^2 to 44 KW/m^2. For sizing an engine, should I just assume the thrust will have to be decreased in order  to accommodate de-icing?

De-Ice is a pretty big topic and with the additions you have made it may warrant its own question all together (One id be happy to work on an answer for). You may want to give some more details on the airframe specs as De-Ice systems vary based on many factors.

This depends heavily on how the plane is set up especially these days with the prevalence of glass in the cockpit even in GA planes. For a point of reference most small GA planes like the Piper Cherokee run 60-100 amp alternators which usually have a bit of headroom in them but power everything and charge the battery just fine. If you want to fly in most airspace you are going to need to power at least a transponder and a radio (more than likely two radios).
The Garmin GTX330 is a 250 Watt transponder and is fairly common in the Archers I fly.
Your run of the mill Garmin GTR 225 radio comes in 10 and 16 watt versions.
To get the total requirement just add up all of your components draw. Most manufacturers list power requirements (or voltage and current draw requirements) for any given piece of equipment and you will have your required power.
As for De-Ice requirements this depends on the type of system you implement. A weeping wing will only require power to drive the fluid pump and the windshield heater where as a boot style system will require enough power to drive the compressor that inflates the boots. Although its becoming increasingly more common to see FIKI on GA planes the majority of the small aircraft fleet still lacks FIKI capability.

The GTX330 has a 250 W RF output power, but it transmits a very small fraction of the time. Consequently, its power draw is 27 watts. Conversely, voice radios are used with long transmit periods, and are never 100% efficient, so their design power draw may be several times RF output.

To follow on from @user71659 - the GTR225 (10W RF) is specified for 8.3W (0.59A @ 14V) of supply power draw (typical) with a maximum operating draw of 60W (4.2A @ 14V), so in terms of the electrical system design you have to build to the maximum expected draw.  There's no getting around the need to crack open the manual for every single instrument or system being installed.